{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "bb425bf5",
   "metadata": {},
   "source": [
    "### Image Math with Qwen3-Omni"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "b92d046f",
   "metadata": {},
   "outputs": [],
   "source": [
    "import os\n",
    "os.environ['VLLM_USE_V1'] = '0'\n",
    "os.environ['VLLM_WORKER_MULTIPROC_METHOD'] = 'spawn'\n",
    "os.environ[\"VLLM_LOGGING_LEVEL\"] = \"ERROR\"\n",
    "os.environ['CUDA_VISIBLE_DEVICES'] = \"0\"\n",
    "import torch\n",
    "import warnings\n",
    "import numpy as np\n",
    "\n",
    "warnings.filterwarnings('ignore')\n",
    "warnings.filterwarnings('ignore', category=DeprecationWarning)\n",
    "warnings.filterwarnings('ignore', category=FutureWarning)\n",
    "warnings.filterwarnings('ignore', category=UserWarning)\n",
    "\n",
    "from qwen_omni_utils import process_mm_info\n",
    "from transformers import Qwen3OmniMoeProcessor\n",
    "\n",
    "def _load_model_processor():\n",
    "    if USE_TRANSFORMERS:\n",
    "        from transformers import Qwen3OmniMoeForConditionalGeneration\n",
    "        if TRANSFORMERS_USE_FLASH_ATTN2:\n",
    "            model = Qwen3OmniMoeForConditionalGeneration.from_pretrained(MODEL_PATH,\n",
    "                                                                         dtype='auto',\n",
    "                                                                         attn_implementation='flash_attention_2',\n",
    "                                                                         device_map=\"auto\")\n",
    "        else:\n",
    "            model = Qwen3OmniMoeForConditionalGeneration.from_pretrained(MODEL_PATH, device_map=\"auto\", dtype='auto')\n",
    "    else:\n",
    "        from vllm import LLM\n",
    "        model = LLM(\n",
    "            model=MODEL_PATH, trust_remote_code=True, gpu_memory_utilization=0.95,\n",
    "            tensor_parallel_size=torch.cuda.device_count(),\n",
    "            limit_mm_per_prompt={'image': 1, 'video': 3, 'audio': 3},\n",
    "            max_num_seqs=1,\n",
    "            max_model_len=32768,\n",
    "            seed=1234,\n",
    "        )\n",
    "\n",
    "    processor = Qwen3OmniMoeProcessor.from_pretrained(MODEL_PATH)\n",
    "    return model, processor\n",
    "\n",
    "def run_model(model, processor, messages, return_audio, use_audio_in_video):\n",
    "    if USE_TRANSFORMERS:\n",
    "        text = processor.apply_chat_template(messages, add_generation_prompt=True, tokenize=False)\n",
    "        audios, images, videos = process_mm_info(messages, use_audio_in_video=use_audio_in_video)\n",
    "        inputs = processor(text=text, audio=audios, images=images, videos=videos, return_tensors=\"pt\", padding=True, use_audio_in_video=use_audio_in_video)\n",
    "        inputs = inputs.to(model.device).to(model.dtype)\n",
    "        text_ids, audio = model.generate(**inputs, \n",
    "                                            thinker_return_dict_in_generate=True,\n",
    "                                            thinker_max_new_tokens=8192, \n",
    "                                            thinker_do_sample=False,\n",
    "                                            speaker=\"Ethan\", \n",
    "                                            use_audio_in_video=use_audio_in_video,\n",
    "                                            return_audio=return_audio)\n",
    "        response = processor.batch_decode(text_ids.sequences[:, inputs[\"input_ids\"].shape[1] :], skip_special_tokens=True, clean_up_tokenization_spaces=False)[0]\n",
    "        if audio is not None:\n",
    "            audio = np.array(audio.reshape(-1).detach().cpu().numpy() * 32767).astype(np.int16)\n",
    "        return response, audio\n",
    "    else:\n",
    "        from vllm import SamplingParams\n",
    "        sampling_params = SamplingParams(temperature=1e-2, top_p=0.1, top_k=1, max_tokens=8192)\n",
    "        text = processor.apply_chat_template(messages, tokenize=False, add_generation_prompt=True)\n",
    "        audios, images, videos = process_mm_info(messages, use_audio_in_video=use_audio_in_video)\n",
    "        inputs = {'prompt': text, 'multi_modal_data': {}, \"mm_processor_kwargs\": {\"use_audio_in_video\": use_audio_in_video}}\n",
    "        if images is not None: inputs['multi_modal_data']['image'] = images\n",
    "        if videos is not None: inputs['multi_modal_data']['video'] = videos\n",
    "        if audios is not None: inputs['multi_modal_data']['audio'] = audios\n",
    "        outputs = model.generate(inputs, sampling_params=sampling_params)\n",
    "        response = outputs[0].outputs[0].text\n",
    "        return response, None\n",
    "    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "d37dcedc",
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "Unrecognized keys in `rope_scaling` for 'rope_type'='default': {'mrope_section', 'interleaved'}\n",
      "`torch_dtype` is deprecated! Use `dtype` instead!\n",
      "2025-09-14 20:45:11,886\tINFO util.py:154 -- Missing packages: ['ipywidgets']. Run `pip install -U ipywidgets`, then restart the notebook server for rich notebook output.\n",
      "The following generation flags are not valid and may be ignored: ['temperature', 'top_p', 'top_k']. Set `TRANSFORMERS_VERBOSITY=info` for more details.\n",
      "You are attempting to use Flash Attention 2 without specifying a torch dtype. This might lead to unexpected behaviour\n",
      "Loading safetensors checkpoint shards:   0% Completed | 0/16 [00:00<?, ?it/s]\n",
      "Loading safetensors checkpoint shards:   6% Completed | 1/16 [00:01<00:15,  1.04s/it]\n",
      "Loading safetensors checkpoint shards:  12% Completed | 2/16 [00:02<00:14,  1.01s/it]\n",
      "Loading safetensors checkpoint shards:  19% Completed | 3/16 [00:03<00:12,  1.00it/s]\n",
      "Loading safetensors checkpoint shards:  25% Completed | 4/16 [00:03<00:11,  1.01it/s]\n",
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      "Loading safetensors checkpoint shards:  38% Completed | 6/16 [00:05<00:09,  1.03it/s]\n",
      "Loading safetensors checkpoint shards:  44% Completed | 7/16 [00:06<00:08,  1.03it/s]\n",
      "Loading safetensors checkpoint shards:  50% Completed | 8/16 [00:07<00:07,  1.02it/s]\n",
      "Loading safetensors checkpoint shards:  56% Completed | 9/16 [00:08<00:06,  1.02it/s]\n",
      "Loading safetensors checkpoint shards:  62% Completed | 10/16 [00:09<00:05,  1.01it/s]\n",
      "Loading safetensors checkpoint shards:  69% Completed | 11/16 [00:10<00:04,  1.02it/s]\n",
      "Loading safetensors checkpoint shards:  75% Completed | 12/16 [00:11<00:03,  1.02it/s]\n",
      "Loading safetensors checkpoint shards:  81% Completed | 13/16 [00:12<00:02,  1.01it/s]\n",
      "Loading safetensors checkpoint shards:  88% Completed | 14/16 [00:13<00:01,  1.01it/s]\n",
      "Loading safetensors checkpoint shards:  94% Completed | 15/16 [00:14<00:00,  1.01it/s]\n",
      "Loading safetensors checkpoint shards: 100% Completed | 16/16 [00:15<00:00,  1.06it/s]\n",
      "Loading safetensors checkpoint shards: 100% Completed | 16/16 [00:15<00:00,  1.02it/s]\n",
      "\n",
      "The image processor of type `Qwen2VLImageProcessor` is now loaded as a fast processor by default, even if the model checkpoint was saved with a slow processor. This is a breaking change and may produce slightly different outputs. To continue using the slow processor, instantiate this class with `use_fast=False`. Note that this behavior will be extended to all models in a future release.\n",
      "Unused or unrecognized kwargs: images.\n",
      "Capturing CUDA graph shapes: 100%|██████████| 1/1 [00:00<00:00,  1.73it/s]\n"
     ]
    }
   ],
   "source": [
    "import librosa\n",
    "import audioread\n",
    "\n",
    "from IPython.display import Video\n",
    "from IPython.display import Image\n",
    "from IPython.display import Audio\n",
    "\n",
    "MODEL_PATH = \"Qwen/Qwen3-Omni-30B-A3B-Thinking\"\n",
    "\n",
    "USE_TRANSFORMERS = False\n",
    "TRANSFORMERS_USE_FLASH_ATTN2 = True\n",
    "\n",
    "model, processor = _load_model_processor()\n",
    "\n",
    "USE_AUDIO_IN_VIDEO = True\n",
    "RETURN_AUDIO = False"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "0b1330dc",
   "metadata": {},
   "outputs": [
    {
     "data": {
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      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {
      "image/jpeg": {
       "height": 360,
       "width": 640
      }
     },
     "output_type": "display_data"
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "Adding requests: 100%|██████████| 1/1 [00:05<00:00,  5.41s/it]\n",
      "Processed prompts: 100%|██████████| 1/1 [01:04<00:00, 64.51s/it, est. speed input: 2.77 toks/s, output: 122.06 toks/s]"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<think>\n",
      "Okay, so I need to figure out the steady rotation rate of this 3-arm lawn sprinkler. Let me start by recalling what I know about sprinklers and fluid mechanics. Hmm, sprinklers are like rotating devices where water flows through the arms and exits, creating a torque that makes the sprinkler spin. Since it's a 3-arm sprinkler, there are three arms, each with a nozzle, right?\n",
      "\n",
      "The problem states that water at 20°C is entering through the center at 2.7 m³/hr. The collar friction is neglected, so we don't have to worry about frictional losses there. We need to find the rotation rate in rev/min when θ is 40 degrees. The figure shows R = 15 cm and d = 7 mm. Let me note down all the given data first.\n",
      "\n",
      "Given:\n",
      "- Flow rate Q = 2.7 m³/hr. Wait, but we need to convert that to m³/s maybe? Because standard units for fluid mechanics are usually SI units. Let me check: 2.7 m³ per hour. So 2.7 divided by 3600 seconds. Let me compute that: 2.7 / 3600 = 0.00075 m³/s. So Q = 0.00075 m³/s.\n",
      "\n",
      "- Temperature 20°C, so water properties: density ρ = 998 kg/m³ (since at 20°C, water density is approximately 998 kg/m³), and viscosity maybe not needed here since it's about momentum.\n",
      "\n",
      "- The sprinkler has three arms, each with a nozzle diameter d = 7 mm. So each nozzle has diameter 7 mm, so radius 3.5 mm. Wait, but how many nozzles? Three arms, so three nozzles. So total flow rate Q is split into three equal parts? Wait, the problem says \"receives 20°C water through the center at 2.7 m³/hr\". So the total flow rate is 2.7 m³/hr, which is divided equally among the three arms? So each arm has Q/3 flow rate.\n",
      "\n",
      "Wait, but maybe not necessarily equal? Hmm, but in a symmetric sprinkler like this, probably each arm has equal flow. So each nozzle has Q/3. Let me confirm that assumption later if needed.\n",
      "\n",
      "- The radius R = 15 cm = 0.15 m. So the distance from the center to the nozzle is 0.15 m.\n",
      "\n",
      "- θ = 40 degrees. Hmm, θ is the angle between the nozzle direction and the radial direction? Wait, looking at the figure description: \"Fig. P3.114\" shows a 3-arm sprinkler with θ as the angle. Probably θ is the angle between the direction of the water jet exiting the nozzle and the tangential direction. Wait, but in sprinkler problems, usually θ is the angle between the nozzle direction and the radial direction. Wait, but depending on the figure. Since the figure is not here, but the problem says \"θ = 40°\", and the options are given. Let me think.\n",
      "\n",
      "In typical sprinkler problems, the angle θ is the angle between the nozzle exit direction and the radial direction. So if θ is 40 degrees, then the water exits at 40 degrees from the radial direction. But depending on the direction, maybe it's the angle between the jet direction and the tangential direction. Hmm. Wait, but in order to get torque, the tangential component of the jet velocity is important. So if the jet is directed at an angle θ from the radial direction, then the tangential component would be related to θ. Alternatively, if θ is measured from the tangential direction, then the radial component would be related. Hmm.\n",
      "\n",
      "Wait, maybe θ is the angle between the jet direction and the direction of rotation. Wait, but the problem states \"θ = 40°\", so probably θ is the angle between the nozzle direction and the radial direction. Let me check standard sprinkler problems.\n",
      "\n",
      "In standard problems, the angle θ is usually the angle between the jet direction and the radial direction. So if θ is 40 degrees, then the jet is directed at 40 degrees from the radial direction. However, depending on the direction of rotation, the angle might be measured differently. Wait, but since it's a steady rotation rate, we need to consider the relative velocity.\n",
      "\n",
      "Wait, another approach: The sprinkler rotates at angular velocity ω. The water exits the nozzle with a velocity relative to the nozzle. But since the nozzle is moving with the sprinkler, the absolute velocity of the water jet is the vector sum of the relative velocity (with respect to the nozzle) and the velocity of the nozzle itself.\n",
      "\n",
      "But in order to find the torque, we need to consider the momentum flux. The torque is equal to the rate of change of angular momentum. So, for each nozzle, the torque contributed is R × (mass flow rate × velocity). But since the sprinkler is rotating, the velocity of the water relative to the ground is different.\n",
      "\n",
      "Wait, let's recall the formula for the torque on a sprinkler. The torque τ is equal to the sum over all nozzles of (mass flow rate per nozzle) × (R × V_tangential), where V_tangential is the tangential component of the absolute velocity of the jet.\n",
      "\n",
      "But the absolute velocity of the jet is the relative velocity (with respect to the nozzle) plus the velocity of the nozzle. Wait, actually, the relative velocity is the velocity of the jet relative to the nozzle. So if the nozzle is moving with tangential velocity ωR, then the absolute velocity of the jet is the relative velocity vector plus the nozzle's velocity vector.\n",
      "\n",
      "Assuming that the water is exiting the nozzle at an angle θ relative to the radial direction. Let's suppose θ is measured from the radial direction towards the tangential direction. So if θ = 0, the jet is radial, and if θ = 90°, it's tangential. But in this problem θ = 40°, so maybe it's 40 degrees from radial towards tangential.\n",
      "\n",
      "But to get the tangential component of the jet velocity, we need to consider the relative velocity. Wait, perhaps the relative velocity is along the nozzle direction. So if the nozzle is angled at θ from radial, then the relative velocity vector is along that direction. Then, the absolute velocity of the jet is the relative velocity plus the tangential velocity of the nozzle.\n",
      "\n",
      "Wait, this is getting confusing. Let me try to recall the standard approach.\n",
      "\n",
      "In sprinkler problems, the torque is calculated by considering the change in angular momentum. The water enters the sprinkler at the center with no angular momentum (since it's axial flow), and exits through the nozzles with angular momentum. The torque required to maintain steady rotation is equal to the rate of change of angular momentum.\n",
      "\n",
      "Assuming that the flow is steady, the torque τ = Σ (m_dot * V_t * R), where V_t is the tangential component of the exit velocity relative to the ground.\n",
      "\n",
      "But the exit velocity relative to the ground is the relative velocity (with respect to the nozzle) plus the velocity of the nozzle. Wait, actually, the relative velocity is the velocity of the jet relative to the nozzle. So if the nozzle is moving with tangential velocity ωR, then the absolute velocity of the jet is the relative velocity vector plus the nozzle's velocity vector.\n",
      "\n",
      "Assuming that the relative velocity is along the nozzle direction. So if the nozzle is angled at θ from the radial direction, then the relative velocity vector has components: radial and tangential. Wait, but depending on the direction of θ.\n",
      "\n",
      "Wait, perhaps θ is the angle between the jet direction and the tangential direction. Hmm. Let's suppose that θ is the angle between the jet direction and the tangential direction. Then, the tangential component of the relative velocity would be V_rel * cosθ, and radial component V_rel * sinθ. But this is speculative.\n",
      "\n",
      "Alternatively, maybe θ is the angle between the jet direction and the radial direction. So if θ = 0, jet is radial; θ = 90°, jet is tangential. Then, the tangential component of the relative velocity would be V_rel * sinθ. Hmm.\n",
      "\n",
      "Wait, perhaps the key is to figure out the tangential component of the exit velocity relative to the ground. Let's denote:\n",
      "\n",
      "Let ω be the angular velocity of the sprinkler (rad/s). Then, the tangential velocity of the nozzle is ωR.\n",
      "\n",
      "The water exits the nozzle with a relative velocity V_rel (relative to the nozzle). If the nozzle is angled at θ from the radial direction, then the relative velocity has components:\n",
      "\n",
      "- Radial component: V_rel * cosθ\n",
      "\n",
      "- Tangential component: V_rel * sinθ\n",
      "\n",
      "But wait, depending on the direction of θ. If θ is measured from the radial direction towards the direction of rotation, then the tangential component would be V_rel * sinθ. However, the absolute tangential velocity of the jet would be the relative tangential component plus the nozzle's tangential velocity. Wait, no: the absolute velocity is the relative velocity plus the velocity of the nozzle.\n",
      "\n",
      "Wait, the nozzle is moving with tangential velocity ωR. So if the relative velocity has a tangential component V_rel_tangential, then the absolute tangential velocity is V_rel_tangential + ωR. But depending on direction. If the jet is directed in the same direction as rotation, then it's added; if opposite, subtracted.\n",
      "\n",
      "Wait, but in a sprinkler, the water is expelled in a direction that creates a torque to rotate the sprinkler. So if the jet is angled such that the tangential component of the relative velocity is opposite to the direction of rotation, then the absolute tangential velocity would be V_rel_tangential - ωR. Hmm, this is getting complicated.\n",
      "\n",
      "Wait, let's consider that the relative velocity V_rel is the velocity of the water exiting the nozzle relative to the nozzle. So if the nozzle is moving with tangential velocity ωR, then the absolute velocity of the water jet is V_rel + ωR (vectorially). But depending on the direction of V_rel.\n",
      "\n",
      "Assuming that θ is the angle between the nozzle direction and the radial direction. So if θ = 0, the jet is radial outward. If θ = 90°, the jet is tangential. But in reality, sprinklers usually have the jet angled so that the tangential component of the relative velocity is in the direction of rotation, causing the sprinkler to spin.\n",
      "\n",
      "Wait, but in order to get a steady rotation rate, the torque due to the jet must balance the torque required to maintain rotation. Wait, but since collar friction is neglected, the only torque is due to the jet. Wait, no: actually, in steady rotation, the torque from the jet must be zero? Wait, no. Wait, if the sprinkler is rotating at a constant speed, then the torque from the jet must be balanced by... but since there's no friction, how does it reach steady rotation? Hmm, maybe the torque from the jet is zero at steady state? Wait, no.\n",
      "\n",
      "Wait, actually, in reality, the torque from the jet causes the sprinkler to accelerate until the torque due to the jet is balanced by the torque required to overcome friction. But since collar friction is neglected, then the sprinkler would accelerate indefinitely? But the problem states \"steady rotation rate\", so maybe there's something I'm missing here.\n",
      "\n",
      "Wait, no, actually, in the case of a sprinkler with no friction, the rotation rate would be such that the tangential component of the jet velocity relative to the ground is zero? Wait, that might be the case. Wait, no. Wait, if the sprinkler is rotating at a certain speed, then the water exiting the nozzle has a relative velocity, but the absolute velocity might have a tangential component. However, if the sprinkler is rotating at a speed where the tangential component of the jet velocity relative to the ground is zero, then there would be no torque. But that doesn't make sense.\n",
      "\n",
      "Wait, perhaps the key is that the torque is equal to the rate of change of angular momentum. Since the water enters with zero angular momentum (axial flow), and exits with angular momentum, the torque τ = Σ (m_dot * V_t * R), where V_t is the tangential component of the exit velocity.\n",
      "\n",
      "But if the sprinkler is rotating at steady state, then the torque must be zero? No, that can't be. Wait, no. Wait, the torque is provided by the water jet, which causes the sprinkler to rotate. However, if there's no friction, then the sprinkler would keep accelerating. But the problem states \"steady rotation rate\", so maybe in this problem, they are considering the case where the torque is balanced by... Hmm, maybe the problem assumes that the water is exiting with a certain velocity relative to the ground, and the torque is calculated based on that. Wait, perhaps the standard approach here is to consider that the relative velocity of the water with respect to the nozzle is such that the tangential component of the absolute velocity is zero? No, that might not be.\n",
      "\n",
      "Wait, let me check the standard formula for sprinkler rotation speed. I recall that for a sprinkler, the angular velocity ω is given by ω = (Q * V_rel * sinθ) / (3 * R * A), but not sure. Wait, maybe better to derive it.\n",
      "\n",
      "Let me consider one nozzle first. Total flow rate Q = 2.7 m³/hr = 0.00075 m³/s. Since there are three arms, each arm has a flow rate of Q/3 = 0.00025 m³/s.\n",
      "\n",
      "The nozzle diameter d = 7 mm = 0.007 m. So area A = π*(d/2)^2 = π*(0.0035)^2 ≈ 3.848e-5 m².\n",
      "\n",
      "Then, the velocity of water relative to the nozzle (V_rel) can be found by Q/3 = A * V_rel => V_rel = (Q/3)/A.\n",
      "\n",
      "Compute V_rel:\n",
      "\n",
      "Q/3 = 0.00025 m³/s\n",
      "\n",
      "A = π*(0.0035)^2 ≈ 3.848e-5 m²\n",
      "\n",
      "V_rel = 0.00025 / 3.848e-5 ≈ 6.496 m/s. Let me compute that:\n",
      "\n",
      "0.00025 / 0.00003848 ≈ 6.496 m/s. Yes.\n",
      "\n",
      "So V_rel ≈ 6.5 m/s.\n",
      "\n",
      "Now, the sprinkler rotates at angular velocity ω. The tangential velocity of the nozzle is ωR. So the absolute tangential velocity of the water jet is V_rel_tangential - ωR (assuming that the jet is directed such that its tangential component relative to the nozzle is in the direction opposite to rotation? Wait, need to be careful with signs.\n",
      "\n",
      "Assuming that θ is the angle between the nozzle direction and the radial direction. So if θ = 40°, then the relative velocity vector has a tangential component of V_rel * sinθ (if θ is measured from radial towards tangential direction). Wait, if θ is measured from radial, then tangential component is V_rel * sinθ, radial component V_rel * cosθ.\n",
      "\n",
      "But depending on the direction of θ. If θ is measured towards the direction of rotation, then the tangential component of the relative velocity would be V_rel * sinθ in the direction of rotation. However, the absolute tangential velocity of the jet would be V_rel * sinθ + ωR? Wait, no.\n",
      "\n",
      "Wait, the nozzle is moving with tangential velocity ωR in the direction of rotation. If the relative velocity has a tangential component V_rel * sinθ in the same direction as rotation, then the absolute tangential velocity would be V_rel * sinθ + ωR. But that would mean the jet is moving faster than the nozzle, which might not be the case.\n",
      "\n",
      "Wait, actually, the relative velocity is the velocity of the jet relative to the nozzle. So if the nozzle is moving at ωR tangentially, then the absolute velocity of the jet is the relative velocity plus the nozzle's velocity.\n",
      "\n",
      "But the direction of the relative velocity: if the nozzle is angled at θ from radial towards the direction of rotation, then the relative velocity has a tangential component V_rel * sinθ in the direction of rotation. Therefore, absolute tangential velocity V_t = V_rel * sinθ + ωR.\n",
      "\n",
      "But wait, if the jet is expelled in the direction of rotation, then the absolute velocity would be higher. However, in reality, for the sprinkler to rotate, the jet must be expelled in a direction that creates a torque. If the jet is expelled in the opposite direction of rotation, then the torque would be in the direction of rotation.\n",
      "\n",
      "Wait, this is confusing. Let's think about it. Suppose the sprinkler is rotating clockwise. If the jet is expelled in a direction that has a tangential component counterclockwise relative to the nozzle, then the absolute tangential velocity would be V_rel_tangential (counterclockwise) plus the nozzle's velocity (clockwise). Depending on which is larger, the absolute velocity could be clockwise or counterclockwise.\n",
      "\n",
      "But in steady rotation, the torque must balance such that the angular momentum change is zero? Wait, no. Wait, torque is equal to the rate of change of angular momentum. Since the sprinkler is rotating at constant speed, the net torque must be zero? But how?\n",
      "\n",
      "Wait, no. Actually, the torque exerted by the water on the sprinkler is equal to the rate of change of angular momentum of the water. Since the water enters with zero angular momentum (axial flow), and exits with angular momentum, the torque τ = Σ (m_dot * V_t * R), where V_t is the tangential component of the exit velocity.\n",
      "\n",
      "But if the sprinkler is rotating at steady state, then this torque must be balanced by... but since there's no friction, how does it reach steady state? Wait, maybe in reality, the torque is zero at steady state? No, that doesn't make sense. Wait, perhaps the problem assumes that the torque is zero because the sprinkler is rotating at a speed where the tangential component of the jet velocity relative to the ground is zero? Wait, that might be the case.\n",
      "\n",
      "Wait, here's a thought: If the sprinkler rotates at a speed such that the tangential component of the jet velocity relative to the ground is zero, then there would be no torque, and thus the sprinkler would rotate at that speed. But why would that be the case? Because if the jet has no tangential component, then there's no change in angular momentum, so torque is zero. But how does the sprinkler reach that speed?\n",
      "\n",
      "Wait, actually, in reality, the sprinkler accelerates until the torque becomes zero. Because as it spins faster, the tangential component of the jet velocity decreases until it becomes zero. At that point, there's no torque, so it stops accelerating. So the steady rotation rate is when the tangential component of the jet velocity relative to the ground is zero. That must be the case here.\n",
      "\n",
      "So, if that's true, then V_t = 0 = V_rel_tangential - ωR. Wait, depending on direction.\n",
      "\n",
      "Assuming that the relative tangential velocity component is in the opposite direction to the rotation. Let me formalize this.\n",
      "\n",
      "Let’s define:\n",
      "\n",
      "- The nozzle is rotating with angular velocity ω (rad/s), so the tangential velocity of the nozzle is ωR.\n",
      "\n",
      "- The water exits the nozzle with a relative velocity V_rel at an angle θ from the radial direction. Let's assume θ is measured from the radial direction towards the direction opposite to rotation. Wait, this is getting too vague.\n",
      "\n",
      "Wait, standard approach: For a sprinkler, the torque is given by τ = m_dot * R * (V_rel * sinθ - ωR), where θ is the angle between the jet direction and the radial direction. Wait, but I need to check.\n",
      "\n",
      "Wait, here's a better way. Let's consider the angular momentum.\n",
      "\n",
      "The angular momentum per unit mass of water exiting the nozzle is R * V_t, where V_t is the tangential component of the absolute velocity.\n",
      "\n",
      "The absolute tangential velocity V_t = V_rel_tangential - ωR. Wait, why minus? Because if the nozzle is moving at ωR in the direction of rotation, and the relative tangential velocity is V_rel_tangential in the opposite direction (to create torque), then absolute velocity would be V_rel_tangential - ωR.\n",
      "\n",
      "Wait, suppose θ is the angle between the jet direction and the radial direction. If θ is measured towards the direction opposite to rotation, then the tangential component of the relative velocity is V_rel * sinθ in the opposite direction. So absolute tangential velocity V_t = -V_rel * sinθ + ωR. Hmm, signs are tricky.\n",
      "\n",
      "Alternatively, let's use the following:\n",
      "\n",
      "Let’s assume that the jet is directed at an angle θ from the radial direction towards the direction of rotation. Then, the relative tangential velocity component is V_rel * sinθ in the direction of rotation. The nozzle is moving at ωR in the direction of rotation. Therefore, the absolute tangential velocity of the jet is V_rel * sinθ + ωR. But this would mean that the jet is moving faster than the nozzle, which would create a torque opposing the rotation? Wait, no.\n",
      "\n",
      "Wait, torque is calculated as the rate of change of angular momentum. The angular momentum of the water exiting the nozzle is R * m_dot * V_t, where V_t is the tangential component of the absolute velocity. Since the water enters with zero angular momentum, the torque τ = Σ (m_dot * R * V_t).\n",
      "\n",
      "But if the sprinkler is rotating at steady state, then the torque must be zero? No, because the torque is what causes the rotation. Wait, no—if there's no friction, then the torque would cause angular acceleration. However, the problem states \"steady rotation rate\", which suggests that the torque is zero. Therefore, the only way for the rotation to be steady is if the torque is zero, which would happen when the tangential component of the jet velocity relative to the ground is zero. So V_t = 0.\n",
      "\n",
      "Therefore, V_t = V_rel_tangential - ωR = 0. Wait, depending on direction.\n",
      "\n",
      "Assuming that the relative tangential velocity component is in the direction opposite to rotation. Let's say θ is the angle between the jet direction and the radial direction, measured towards the direction opposite to rotation. Then, the relative tangential velocity component is V_rel * sinθ opposite to rotation. The nozzle is moving at ωR in the direction of rotation. Therefore, absolute tangential velocity V_t = -V_rel * sinθ + ωR. Setting V_t = 0 gives ωR = V_rel * sinθ => ω = (V_rel * sinθ)/R.\n",
      "\n",
      "Wait, that seems plausible. Let me check.\n",
      "\n",
      "If the jet is expelled in a direction that has a tangential component opposite to the rotation direction, then the absolute tangential velocity would be ωR (nozzle moving in rotation direction) minus V_rel * sinθ (jet moving opposite). If these are equal, then V_t = 0, so no torque. Therefore, the sprinkler rotates at ω = (V_rel * sinθ)/R.\n",
      "\n",
      "Yes, this seems right. So the steady rotation rate occurs when the tangential component of the jet velocity relative to the ground is zero, so that there's no torque. Wait, but why would that be the case? Because if there's no torque, then the sprinkler would maintain constant angular velocity. Since there's no friction, once it reaches that speed, it would keep rotating at that speed. So yes, this must be the condition.\n",
      "\n",
      "Therefore, ω = (V_rel * sinθ)/R.\n",
      "\n",
      "Wait, but let's confirm with the formula.\n",
      "\n",
      "Torque τ = Σ (m_dot * R * V_t), where V_t is the tangential component of the absolute velocity.\n",
      "\n",
      "If V_t = 0, then τ = 0, so no acceleration. Therefore, steady rotation.\n",
      "\n",
      "So V_t = 0 = V_rel_tangential - ωR.\n",
      "\n",
      "Assuming V_rel_tangential is the tangential component of the relative velocity. If θ is measured from radial towards the direction opposite to rotation, then V_rel_tangential = V_rel * sinθ. So:\n",
      "\n",
      "ωR = V_rel * sinθ => ω = (V_rel * sinθ)/R.\n",
      "\n",
      "Yes, that seems correct.\n",
      "\n",
      "So let's compute this.\n",
      "\n",
      "First, compute V_rel.\n",
      "\n",
      "Given:\n",
      "\n",
      "Total flow rate Q = 2.7 m³/hr = 2.7 / 3600 = 0.00075 m³/s.\n",
      "\n",
      "Each nozzle has Q/3 = 0.00025 m³/s.\n",
      "\n",
      "Nozzle diameter d = 7 mm = 0.007 m.\n",
      "\n",
      "Area A = π*(d/2)^2 = π*(0.0035)^2 ≈ 3.848e-5 m².\n",
      "\n",
      "V_rel = (Q/3)/A = 0.00025 / 3.848e-5 ≈ 6.496 m/s.\n",
      "\n",
      "θ = 40°, so sinθ = sin(40°) ≈ 0.6428.\n",
      "\n",
      "R = 15 cm = 0.15 m.\n",
      "\n",
      "Thus,\n",
      "\n",
      "ω = (V_rel * sinθ)/R = (6.496 * 0.6428) / 0.15 ≈ (4.176) / 0.15 ≈ 27.84 rad/s.\n",
      "\n",
      "Convert rad/s to rev/min:\n",
      "\n",
      "1 rev = 2π rad, so ω (rev/s) = 27.84 / (2π) ≈ 4.43 rev/s.\n",
      "\n",
      "Multiply by 60: 4.43 * 60 ≈ 265.8 rev/min.\n",
      "\n",
      "Wait, but the options are A: 317, B: 414, C: 400, D: NaN.\n",
      "\n",
      "Hmm, 265 is not among the options. So something is wrong here.\n",
      "\n",
      "Wait, maybe my assumption about θ is incorrect. Maybe θ is measured from the tangential direction instead of radial.\n",
      "\n",
      "Let me check. If θ is the angle from the tangential direction, then the radial component would be V_rel * sinθ, and tangential component V_rel * cosθ.\n",
      "\n",
      "Wait, if θ is measured from tangential towards radial, then the tangential component of relative velocity is V_rel * cosθ.\n",
      "\n",
      "But then, if the jet is directed at θ from tangential towards radial, then the tangential component of relative velocity is V_rel * cosθ.\n",
      "\n",
      "Assuming that the relative tangential velocity component is in the direction opposite to rotation, then:\n",
      "\n",
      "ωR = V_rel * cosθ => ω = (V_rel * cosθ)/R.\n",
      "\n",
      "Let's compute that.\n",
      "\n",
      "cos(40°) ≈ 0.7660.\n",
      "\n",
      "So ω = (6.496 * 0.7660)/0.15 ≈ (4.975)/0.15 ≈ 33.17 rad/s.\n",
      "\n",
      "Convert to rev/min: 33.17 / (2π) * 60 ≈ (5.278) * 60 ≈ 316.7 rev/min ≈ 317 rev/min.\n",
      "\n",
      "Ah! That's option A.\n",
      "\n",
      "So probably θ is measured from the tangential direction. So if θ = 40°, then the angle between the jet direction and the tangential direction is 40°, so the tangential component of the relative velocity is V_rel * cosθ.\n",
      "\n",
      "Wait, but why would θ be measured from tangential? Let me think.\n",
      "\n",
      "In some textbooks, θ is defined as the angle between the jet direction and the tangential direction. So if θ = 0, the jet is tangential; θ = 90°, radial.\n",
      "\n",
      "Given that the answer matches option A when θ is measured from tangential, that must be the case.\n",
      "\n",
      "So let's recap:\n",
      "\n",
      "If θ is the angle between the jet direction and the tangential direction, then the tangential component of the relative velocity is V_rel * cosθ.\n",
      "\n",
      "Assuming that the jet is directed such that the tangential component of the relative velocity is opposite to the direction of rotation (so that the absolute tangential velocity is zero when ωR = V_rel * cosθ), then ω = (V_rel * cosθ)/R.\n",
      "\n",
      "Wait, but why opposite? Let's think.\n",
      "\n",
      "If the sprinkler is rotating clockwise, then the nozzle is moving clockwise. If the jet is expelled in a direction that has a tangential component counterclockwise (relative to the nozzle), then the absolute tangential velocity would be V_rel_tangential (counterclockwise) + ωR (clockwise). Wait, signs are confusing.\n",
      "\n",
      "Alternatively, using the formula τ = m_dot * R * (V_rel_tangential - ωR). For steady rotation, τ = 0, so V_rel_tangential = ωR.\n",
      "\n",
      "But V_rel_tangential is the tangential component of the relative velocity. If θ is measured from tangential, then V_rel_tangential = V_rel * cosθ.\n",
      "\n",
      "Wait, if θ is the angle between the jet direction and the tangential direction, then yes, the tangential component is V_rel * cosθ.\n",
      "\n",
      "Assuming that the jet is expelled in the direction opposite to rotation, so that V_rel_tangential is negative, but magnitude-wise, ωR = V_rel * cosθ.\n",
      "\n",
      "Wait, regardless of direction, the magnitude would be ω = (V_rel * cosθ)/R.\n",
      "\n",
      "So with θ = 40°, cosθ ≈ 0.766, so ω = (6.496 * 0.766)/0.15 ≈ 33.17 rad/s ≈ 317 rev/min.\n",
      "\n",
      "Yes, that's option A.\n",
      "\n",
      "But why did I get confused earlier? Because of the definition of θ. Since the answer matches option A when θ is measured from tangential, that must be the correct interpretation.\n",
      "\n",
      "Let me verify the calculation again step by step.\n",
      "\n",
      "Given:\n",
      "\n",
      "Q = 2.7 m³/hr = 2.7 / 3600 = 0.00075 m³/s.\n",
      "\n",
      "Each nozzle: Q/3 = 0.00025 m³/s.\n",
      "\n",
      "d = 7 mm = 0.007 m.\n",
      "\n",
      "Area A = π*(0.007/2)^2 = π*(0.0035)^2 ≈ 3.848e-5 m².\n",
      "\n",
      "V_rel = (Q/3)/A = 0.00025 / 3.848e-5 ≈ 6.496 m/s.\n",
      "\n",
      "θ = 40°, so cosθ = cos(40°) ≈ 0.7660.\n",
      "\n",
      "R = 0.15 m.\n",
      "\n",
      "ω = (V_rel * cosθ)/R = (6.496 * 0.7660)/0.15 ≈ (4.975)/0.15 ≈ 33.17 rad/s.\n",
      "\n",
      "Convert to rev/min:\n",
      "\n",
      "33.17 rad/s * (1 rev / 2π rad) * 60 s/min ≈ 33.17 * 60 / (2 * 3.1416) ≈ 33.17 * 9.549 ≈ 316.7 rev/min ≈ 317 rev/min.\n",
      "\n",
      "Yes, that's exactly option A.\n",
      "\n",
      "Therefore, the correct answer is A, 317 rev/min.\n",
      "\n",
      "**Final Answer**\n",
      "\\boxed{A}\n",
      "<\n",
      "</think>\n",
      "\n",
      "To determine the steady rotation rate of the 3-arm lawn sprinkler, we analyze the fluid dynamics and angular momentum principles involved in such a system.\n",
      "\n",
      "---\n",
      "\n",
      "### **Given Data:**\n",
      "\n",
      "- **Total flow rate:**  \n",
      "  $ Q = 2.7 \\, \\text{m}^3/\\text{hr} = \\frac{2.7}{3600} = 0.00075 \\, \\text{m}^3/\\text{s} $\n",
      "\n",
      "- **Number of arms (nozzles):**  \n",
      "  $ n = 3 $\n",
      "\n",
      "- **Flow rate per nozzle:**  \n",
      "  $ Q_{\\text{nozzle}} = \\frac{Q}{3} = \\frac{0.00075}{3} = 0.00025 \\, \\text{m}^3/\\text{s} $\n",
      "\n",
      "- **Nozzle diameter:**  \n",
      "  $ d = 7 \\, \\text{mm} = 0.007 \\, \\text{m} $\n",
      "\n",
      "- **Radius from center to nozzle:**  \n",
      "  $ R = 15 \\, \\text{cm} = 0.15 \\, \\text{m} $\n",
      "\n",
      "- **Angle between jet direction and tangential direction:**  \n",
      "  $ \\theta = 40^\\circ $\n",
      "\n",
      "- **Water density at 20°C:**  \n",
      "  $ \\rho = 998 \\, \\text{kg/m}^3 $ (not needed for this calculation)\n",
      "\n",
      "---\n",
      "\n",
      "### **Step 1: Compute the Relative Velocity of Water Exiting the Nozzle**\n",
      "\n",
      "The velocity of water relative to the nozzle is:\n",
      "\n",
      "$$\n",
      "V_{\\text{rel}} = \\frac{Q_{\\text{nozzle}}}{A}\n",
      "$$\n",
      "\n",
      "Where the cross-sectional area of the nozzle is:\n",
      "\n",
      "$$\n",
      "A = \\pi \\left( \\frac{d}{2} \\right)^2 = \\pi \\left( \\frac{0.007}{2} \\right)^2 = \\pi (0.0035)^2 \\approx 3.848 \\times 10^{-5} \\, \\text{m}^2\n",
      "$$\n",
      "\n",
      "$$\n",
      "V_{\\text{rel}} = \\frac{0.00025}{3.848 \\times 10^{-5}} \\approx 6.496 \\, \\text{m/s}\n",
      "$$\n",
      "\n",
      "---\n",
      "\n",
      "### **Step 2: Determine the Tangential Component of Relative Velocity**\n",
      "\n",
      "Assuming $ \\theta $ is the angle between the jet direction and the **tangential direction**, the tangential component of the relative velocity is:\n",
      "\n",
      "$$\n",
      "V_{\\text{rel,tang}} = V_{\\text{rel}} \\cdot \\cos(\\theta)\n",
      "$$\n",
      "\n",
      "$$\n",
      "\\cos(40^\\circ) \\approx 0.7660\n",
      "$$\n",
      "\n",
      "$$\n",
      "V_{\\text{rel,tang}} = 6.496 \\times 0.7660 \\approx 4.975 \\, \\text{m/s}\n",
      "$$\n",
      "\n",
      "---\n",
      "\n",
      "### **Step 3: Apply the Steady Rotation Condition**\n",
      "\n",
      "At steady rotation, the **tangential component of the absolute velocity of the jet is zero**, which implies:\n",
      "\n",
      "$$\n",
      "\\omega R = V_{\\text{rel,tang}}\n",
      "$$\n",
      "\n",
      "$$\n",
      "\\omega = \\frac{V_{\\text{rel,tang}}}{R} = \\frac{4.975}{0.15} \\approx 33.17 \\, \\text{rad/s}\n",
      "$$\n",
      "\n",
      "---\n",
      "\n",
      "### **Step 4: Convert Angular Velocity to Revolutions per Minute**\n",
      "\n",
      "$$\n",
      "\\omega_{\\text{rev/min}} = \\frac{33.17}{2\\pi} \\times 60 \\approx \\frac{33.17}{6.283} \\times 60 \\approx 5.278 \\times 60 \\approx 316.7 \\, \\text{rev/min}\n",
      "$$\n",
      "\n",
      "Rounding to the nearest whole number:\n",
      "\n",
      "$$\n",
      "\\omega_{\\text{rev/min}} \\approx 317 \\, \\text{rev/min}\n",
      "$$\n",
      "\n",
      "---\n",
      "\n",
      "### **Conclusion**\n",
      "\n",
      "The steady rotation rate of the sprinkler is:\n",
      "\n",
      "$$\n",
      "\\boxed{A}\n",
      "$$\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "\n"
     ]
    }
   ],
   "source": [
    "image_path = \"https://qianwen-res.oss-cn-beijing.aliyuncs.com/Qwen3-Omni/cookbook/5195.jpg\"\n",
    "\n",
    "messages = [\n",
    "    {\n",
    "        \"role\": \"user\",\n",
    "        \"content\": [\n",
    "            {\"type\": \"text\", \"text\": \"The 3-arm lawn sprinkler of Fig. P3.114 receives 20°C water through the center at 2.7 m^3/hr. If collar friction is neglected, what is the steady rotation rate in rev/min for $\\\\theta $ = 40°?\"},\n",
    "            {\"type\": \"image\", \"image\": image_path},\n",
    "            {\"type\": \"text\", \"text\": \"A, 317 rev/min, B, 414 rev/min, C, 400 rev/min, D, NaN, choose one of the options.\"\"\"} \n",
    "        ]\n",
    "    }\n",
    "]\n",
    "\n",
    "display(Image(image_path, width=640, height=360))\n",
    "\n",
    "response, _ = run_model(model=model, messages=messages, processor=processor, return_audio=RETURN_AUDIO, use_audio_in_video=USE_AUDIO_IN_VIDEO)\n",
    "\n",
    "print(response)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "61c7f53d",
   "metadata": {},
   "outputs": [
    {
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      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {
      "image/jpeg": {
       "height": 360,
       "width": 640
      }
     },
     "output_type": "display_data"
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "Adding requests: 100%|██████████| 1/1 [00:00<00:00, 178.79it/s]\n",
      "Processed prompts: 100%|██████████| 1/1 [00:08<00:00,  8.54s/it, est. speed input: 42.74 toks/s, output: 123.54 toks/s]"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<think>\n",
      "Got it, let's try to solve this problem step by step. First, let's recall what the cost function is. The cost function J(θ₁) is given by (1/(2m)) times the sum of squared differences between the hypothesis h_θ(x^i) and the actual y^i for each example. Here, m is 3, so we need to find the three data points from the graph, compute h_θ(x^i) when θ₁ is 0, then compute the squared errors, sum them, and multiply by 1/(2*3).\n",
      "\n",
      "First, let's identify the data points. The graph has x on the horizontal axis and h(x) (which is y) on the vertical axis. Let's list the (x, y) pairs:\n",
      "\n",
      "- The first 'X' is at x=1, y=1. So (x¹, y¹) = (1, 1)\n",
      "- The second 'X' is at x=2, y=2. So (x², y²) = (2, 2)\n",
      "- The third 'X' is at x=3, y=3. So (x³, y³) = (3, 3)\n",
      "\n",
      "Wait, let's check the graph again. The x-axis goes from 0 to 3, with marks at 0,1,2,3. The y-axis is h(x), with marks at 0,1,2,3. The X's are at:\n",
      "\n",
      "- One X at x=1, y=1 (since it's on the vertical line x=1, and the horizontal line y=1)\n",
      "- One X at x=2, y=2 (x=2, y=2)\n",
      "- One X at x=3, y=3 (x=3, y=3)\n",
      "\n",
      "Yes, that's correct. So the three examples are (1,1), (2,2), (3,3).\n",
      "\n",
      "Now, the hypothesis is h_θ(x) = θ₁x. When θ₁ = 0, h_θ(x) = 0*x = 0 for all x. So for each example, h_θ(x^i) = 0.\n",
      "\n",
      "Now, compute the squared error for each example:\n",
      "\n",
      "- For i=1: (h_θ(x¹) - y¹)² = (0 - 1)² = 1\n",
      "- For i=2: (0 - 2)² = 4\n",
      "- For i=3: (0 - 3)² = 9\n",
      "\n",
      "Sum of squared errors: 1 + 4 + 9 = 14\n",
      "\n",
      "Now, the cost function J(θ₁) is (1/(2m)) times this sum. Here, m=3, so 2m=6. Therefore, J(0) = 14 / 6.\n",
      "\n",
      "Let's check the options. Option D is 14/6. So that's the answer.\n",
      "</think>\n",
      "\n",
      "To solve for \\( J(0) \\), we follow these steps:  \n",
      "\n",
      "\n",
      "### Step 1: Identify the training examples  \n",
      "From the graph, the three data points \\((x^i, y^i)\\) are:  \n",
      "- \\((x^1, y^1) = (1, 1)\\)  \n",
      "- \\((x^2, y^2) = (2, 2)\\)  \n",
      "- \\((x^3, y^3) = (3, 3)\\)  \n",
      "\n",
      "\n",
      "### Step 2: Evaluate the hypothesis \\( h_\\theta(x) \\) at \\( \\theta_1 = 0 \\)  \n",
      "The hypothesis is \\( h_\\theta(x) = \\theta_1 x \\). When \\( \\theta_1 = 0 \\), \\( h_\\theta(x) = 0 \\cdot x = 0 \\) for all \\( x \\).  \n",
      "\n",
      "\n",
      "### Step 3: Compute the squared errors  \n",
      "For each example, calculate \\( (h_\\theta(x^i) - y^i)^2 \\):  \n",
      "- For \\( i=1 \\): \\( (0 - 1)^2 = 1 \\)  \n",
      "- For \\( i=2 \\): \\( (0 - 2)^2 = 4 \\)  \n",
      "- For \\( i=3 \\): \\( (0 - 3)^2 = 9 \\)  \n",
      "\n",
      "Sum of squared errors: \\( 1 + 4 + 9 = 14 \\).  \n",
      "\n",
      "\n",
      "### Step 4: Apply the cost function formula  \n",
      "The cost function is \\( J(\\theta_1) = \\frac{1}{2m} \\sum_{i=1}^m (h_\\theta(x^i) - y^i)^2 \\). Here, \\( m = 3 \\), so:  \n",
      "\\[\n",
      "J(0) = \\frac{1}{2 \\cdot 3} \\cdot 14 = \\frac{14}{6}\n",
      "\\]  \n",
      "\n",
      "\n",
      "Thus, the correct answer is **D**.\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "\n"
     ]
    }
   ],
   "source": [
    "image_path = \"https://qianwen-res.oss-cn-beijing.aliyuncs.com/Qwen3-Omni/cookbook/4181.jpg\"\n",
    "\n",
    "messages = [\n",
    "    {\n",
    "        \"role\": \"user\",\n",
    "        \"content\": [\n",
    "            {\"type\": \"text\", \"text\": \"Suppose we have a training set with m=3 examples,plotted below.Our hypothes is representation is $h_{\\\\theta}(x) = \\\\theta_1x$, with parameter $\\\\theta_1$. The cost function $J(\\\\theta_1)$ is $J(\\\\theta_1)$=${1\\\\over2m}\\\\sum\\\\nolimits_{i=1}^m(h_\\\\theta(x^i) - y^i)^2$. What is $J(0)$?\"},\n",
    "            {\"type\": \"image\", \"image\": image_path},\n",
    "            {\"type\": \"text\", \"text\": \"Options:\\nA. 0\\nB. 1/6\\nC. 1\\nD. 14/6\\nPlease select the correct answer from the options above.\"}\n",
    "        ]\n",
    "    }\n",
    "]\n",
    "\n",
    "display(Image(image_path, width=640, height=360))\n",
    "\n",
    "response, _ = run_model(model=model, messages=messages, processor=processor, return_audio=RETURN_AUDIO, use_audio_in_video=USE_AUDIO_IN_VIDEO)\n",
    "\n",
    "print(response)"
   ]
  }
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